A car starts from rest and accelerates at a constant rate in a straight line. In

Question

A car starts from rest and accelerates at a constant rate in a straight line. In the first second the car covers a distance of 2.0 meters. How much additional distance will the car cover during the second second of its motion? 2.0 m 4.0 m 6 0 m 8.0m 13.0 m A car starts from rest and accelerates at a constant rate in a straight line. In the first second the car covers a distance of 2.0 meters. How fast will the car be moving at the end of the second second? 16m/s 4.0 m/s 2.0m/s 8.0m/s 32.0 m/s A brick is dropped from rest from a height of 4.9 m. How long does it take for the brick to reach the ground? 0.6 s 1.0 s 1.2 s 1.4s 2.0s A ball is shot straight up from the surface of the earth with an initial speed of 19.6 m/s. Neglect any effects due to air resistance. What is the magnitude of the ball is displacement from the starting point after 1 00 second has elapsed? 9.80 m 14.7 m 19.6 m 24.5 m 58.8 m What maximum height will the ball reach? 9.80 m 14.7 m 19.6 m 24.5 m 58.8 m How much time elapses between the throwing of the ball and its return to the original launch point? 4.00 s 2.00 s 12.0 s 8.0s 16.0 s A physics student standing on the edge of a cliff throws a stone vertically downward with an initial speed of 10.0 m/s. The instant before the stone hits the ground below, it is traveling with a speed of 30.0 m/s. If the physics student were to throw the rock horizontally outward from the cliff instead, with the same initial speed of 10 m/s, how fast would the stone be traveling just before it hits the ground? 10 m/s 20 m/s 30 m/s 40 m/s The height of the cliff must be specified to answer this question

Explanation / Answer

here,

1)

for t = 1 s

s = 2 m

let the accelration be a

s = u * t + 0.5 * a * t^2

2 = 0 + 0.5 * a * 1^2

a = 4 m/s^2

final velocity , v = u + a * t

v = 4 m/s

for seccond second

s2 = u2*t + 0.5 * a * t^2

s2 = 4 * 1 + 0.5 * 4 * 1^2

s2 = 6 m

the distance covered in seccond second is (3) 6 m

2)

final speed after seccond second , v2 = v1 + a * 1

v2 = 4 + 4 = 8m/s

the final speed is (d) 8 m/s

3)

height of brick , h = 4.9 m

let the time taken be t

h = u * t + 0.5* a*t^2

4.9 = 0 + 0.5 * 9.8 * t^2

t = 1 s

the time taken is (b) 1 s

4)

initial speed , u = 19.6 m/s

h = u* t - 0.5 * g * t^2

h = 19.6 * 1 - 0.5 * 9.8 * 1^2

h = 14.7 m

the height achived is (b) 14.7 m

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